Find the Magnification of the Image. Again

The Mirror Equation - Concave Mirrors

Ray diagrams can be used to determine the image location, size, orientation and blazon of image formed of objects when placed at a given location in front of a concave mirror. The utilise of these diagrams was demonstrated earlier in Lesson 3. Ray diagrams provide useful data about object-image relationships, yet fail to provide the information in a quantitative form. While a ray diagram may help one decide the gauge location and size of the image, information technology will not provide numerical information virtually epitome altitude and object size. To obtain this type of numerical information, information technology is necessary to utilise the Mirror Equation and the Magnification Equation . The mirror equation expresses the quantitative relationship betwixt the object distance (d_o), the image distance (d_i), and the focal length (f). The equation is stated as follows:

The magnification equation relates the ratio of the image altitude and object distance to the ratio of the image elevation (h_i) and object top (h_o). The magnification equation is stated as follows:

These two equations tin can be combined to yield data about the prototype distance and epitome height if the object distance, object height, and focal length are known.

Every bit a sit-in of the effectiveness of the mirror equation and magnification equation, consider the following example problem and its solution.

Example Problem #1

A four.00-cm tall low-cal bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of fifteen.ii cm. Decide the image distance and the image size.

Like all problems in physics, brainstorm by the identification of the known information.

ho = four.0 cm

do = 45.7 cm

f = 15.2 cm

Next place the unknown quantities that you wish to solve for.

To determine the image distance, the mirror equation must exist used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

1/f = 1/practice + 1/d_i

ane/(15.2 cm) = 1/(45.7 cm) + ane/d_i

0.0658 cm^-ane = 0.0219 cm^-1 + 1/d_i

0.0439 cm^-one = 1/d_i

The numerical values in the solution higher up were rounded when written down, yet united nations-rounded numbers were used in all calculations. The final answer is rounded to the 3rd significant digit.

To decide the image meridian, the magnification equation is needed. Since 3 of the four quantities in the equation (disregarding the G) are known, the quaternary quantity can be calculated. The solution is shown below.

h_i/h_o = - d_i/d_o

h_i /(4.0 cm) = - (22.eight cm)/(45.vii cm)

h_i = - (4.0 cm) • (22.8 cm)/(45.7 cm)

The negative values for prototype height indicate that the image is an inverted epitome. As is often the case in physics, a negative or positive sign in front of the numerical value for a physical quantity represents data about management. In the case of the prototype height, a negative value e'er indicates an inverted image.

From the calculations in this problem it tin exist concluded that if a four.00-cm alpine object is placed 45.vii cm from a concave mirror having a focal length of 15.2 cm, then the epitome volition be inverted, ane.99-cm tall and located 22.8 cm from the mirror. The results of this calculation agree with the principles discussed earlier in this lesson. In this example, the object is located across the center of curvature (which would be two focal lengths from the mirror), and the paradigm is located betwixt the center of curvature and the focal indicate. This falls into the category of Instance 1 : The object is located across C.

 

Now lets effort a second case problem:

Example Trouble #2

A 4.0-cm tall low-cal seedling is placed a altitude of 8.3 cm from a concave mirror having a focal length of 15.2 cm. (NOTE: this is the aforementioned object and the same mirror, merely this fourth dimension the object is placed closer to the mirror.) Determine the image distance and the image size.

Again, begin past the identification of the known data.

ho = 4.0 cm

do = 8.3 cm

f = 15.2 cm

Side by side place the unknown quantities that yous wish to solve for.

To determine the epitome altitude, the mirror equation will have to be used. The post-obit lines represent the solution to the prototype distance; substitutions and algebraic steps are shown.

1/f = ane/practice + i/d_i

i/(fifteen.two cm) = 1/(8.3 cm) + ane/d_i

0.0658 cm^-1 = 0.120 cm^-ane + 1/d_i

-0.0547 cm^-i = i/d_i

The numerical values in the solution above were rounded when written downward, yet united nations-rounded numbers were used in all calculations. The last answer is rounded to the third significant digit.

To determine the image height, the magnification equation is needed. Since three of the iv quantities in the equation (disregarding the Thou) are known, the fourth quantity can be calculated. The solution is shown below.

h_i/h_o = - d_i/d_o

h_i /(iv.0 cm) = - (-18.2 cm)/(eight.iii cm)

h_i = - (4.0 cm) • (-xviii.two cm)/(8.3 cm)

The negative value for prototype distance indicates that the prototype is a virtual image located behind the mirror. Again, a negative or positive sign in front end of the numerical value for a concrete quantity represents information nigh management. In the case of the image distance, a negative value always means backside the mirror. Note also that the image height is a positive value, meaning an upright image. Any image that is upright and located behind the mirror is considered to be a virtual image.

From the calculations in the second example problem it can be concluded that if a 4.0-cm tall object is placed 8.three cm from a concave mirror having a focal length of 15.two cm, then the image will be magnified, upright, eight.8-cm tall and located 18.iii cm behind the mirror. The results of this calculation agree with the principles discussed before in this lesson. In this case, the object is located in front of the focal point (i.eastward., the object distance is less than the focal length), and the image is located behind the mirror. This falls into the category of Instance 5: The object is located in front of F.

The +/- Sign Conventions

The sign conventions for the given quantities in the mirror equation and magnification equations are as follows:

• f is + if the mirror is a concave mirror
• f is - if the mirror is a convex mirror
• d_i is + if the image is a existent image and located on the object's side of the mirror.
• d_i is - if the epitome is a virtual epitome and located behind the mirror.
• h_i is + if the image is an upright image (and therefore, too virtual)
• h_i is - if the image an inverted prototype (and therefore, also real)

Like many mathematical problems in physics, the skill is only acquired through much personal do. Perhaps you would like to have some time to try the bug in the Bank check Your Understanding section below.

We Would Like to Propose ...

Why only read about it and when you could be interacting with information technology? Interact - that'due south exactly what you practice when you lot use 1 of The Physics Classroom's Interactives. We would like to suggest that you lot combine the reading of this page with the apply of our Eyes Demote Interactive or our Name That Image Interactive. You can find this in the Physics Interactives section of our website. The Eyes Bench Interactive provides the learner an interactive enivronment for exploring the formation of images past lenses and mirrors. The Proper name That Image Interactive provides learners with an intensive mental workout in recognizing the image characteristics for whatsoever given object location in front end of a curved mirror.

Check Your Understanding

i. Determine the prototype distance and image tiptop for a v.00-cm alpine object placed 45.0 cm from a concave mirror having a focal length of 15.0 cm.

2. Determine the prototype distance and image height for a v.00-cm alpine object placed xxx.0 cm from a concave mirror having a focal length of 15.0 cm.

3. Determine the epitome distance and prototype pinnacle for a 5.00-cm tall object placed 20.0 cm from a concave mirror having a focal length of 15.0 cm.

iv. Determine the image distance and image height for a 5.00-cm tall object placed x.0 cm from a concave mirror having a focal length of 15.0 cm.

5. A magnified, inverted epitome is located a altitude of 32.0 cm from a concave mirror with a focal length of 12.0 cm. Determine the object distance and tell whether the image is real or virtual.

ZINGER : 6. An inverted image is magnified by 2 when the object is placed 22 cm in front of a concave mirror. Determine the image distance and the focal length of the mirror.

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Source: https://www.physicsclassroom.com/class/refln/Lesson-3/The-Mirror-Equation

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